Related : How do I show that for linearly independent set in dual is a dual of a linearly independent set?
Let $V$ be a real normed space and $L(V,\mathbb{R})$ be the space of continuous linear functionals.
Let $\{\lambda_1,...,\lambda_n\}$ and $\{\mu_1,...,\mu_n\}$ be linearly independent subsets of $ L(V,\mathbb{R})$.
Now, define $V_1:=\{x\in V: \forall 1\leq i \leq n, \lambda_i(x)\geq 0\}$ and $V_2:=\{x\in V:\forall 1\leq i \leq n, \mu_i(x)\geq 0\}$
Then, how do I prove that there is a linear homeomorphim $T:V\rightarrow V$ such that $T(V_1)=V_2$?
Here is how I tried. It is proven in the answer in the link above that the map $\phi:V\rightarrow \mathbb{R}^n:x\mapsto (\lambda_1(x),...,\lambda_n(x))$ is surjective and $\psi:V\rightarrow \mathbb{R}^n:x\mapsto (\mu_1(x),...,\mu_n(x))$ is surjective.
Hence, there exists $x_1,...,x_n,y_1,...,y_n\in V$ such that $\lambda_i(x_j)=\mu_i(y_j)=\delta_{ij}$. Thus, $V=\ker(\phi)\oplus span(\{x_1,....x_n\}=\ker(\psi)\oplus span(\{y_1,...,y_n\})$. I tried to construct a linear homeomorphism between the kernels of $\phi$ and $\psi$ then extend it to a linear homeomorphism between $V$ and $V$, but I could not since $V$ may be infinite-dimensional.
How do I prove this? Thank you in advance.